Question 164862
Let x be the length of the side of the square1,
p be the perimeter of the square 1
and a be the area of the square 1.
Then p=4x and a=x^2

Let y be the length of the side of the square2,
q be the perimeter of the square2
and b be the area of the square2.
Then q=4y and b=y^2

Given that the perimeter of one square exceeds that of another by 16 and its area is 44 less than 4 times the area of the other
that is p=q+16 and a=4b-44
that is 4x=4y+16 and x^2=4y^2-44
Now solving these two equations for x and y we get
plug x=y+4  in x^2=4y^2-44
that gives (y+4)^2=4y^2-44
y^2+8y+16=4y^2-44
3y^2-8y-60=0
(3y+10)(y-6)=0
that is y=-10/3 or y=6
since y is the length of the side of the square2 it cant be negative so -10/3 is discarded
hence y=6
therefore x=y+4=6+4=10

Thus length of the side of square1 is 4units and length of the side of the square2 is 10units.

My suggestion is not to worry that the answer is too length try to go through it twice or thrice so that you will understand it more explicitly

I hope you will enjoy it.